Question

At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the left side of the pivot an adult pushes straight down on the teeter-totter with a force of 151 N.In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3.0 m from the pivot? (clockwise/ counterclockwise)

Answer 1

The teeter-totter rotates counterclockwise

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Further explanation

Let's recall Moment of Force as follows:

$\boxed {\tau = F d }$

where:

τ = moment of force ( Nm )

F = magnitude of force ( N )

d = perpendicular distance between force and pivot ( m )

Let us now tackle the problem !

Given:

mass of child = m_c = 21 kg

distance between pivot and child = d_c = 1.8 m

magnitude of force = F = 151 N

distance between force and pivot = d = 3.0 m

Asked:

direction of rotation = ?

Solution:

Firstly , we will find clockwise moment by weight of the child:

$\tau_{clockwise} = w \times d_c$

$\tau_{clockwise} = mg \times d_c$

$\tau_{clockwise} = ( 21 \times 9.8 ) \times 1.8$

$\boxed{\tau_{clockwise} = 370.44 \texttt{ Nm}}$

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Next, we will find counterclockwise moment by pushing force:

$\tau_{counterclockwise} = F \times d$

$\tau_{counterclockwise} = 151 \times 3.0$

$\boxed{\tau_{counterclockwise} = 453 \texttt{ Nm}}$

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Conclusion:

$\tau_{counterclockwise} > \tau_{clockwise}$

$\texttt{The teeter-totter rotates \boxed{counterclockwise}}$

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Answer details

Grade: High School

Subject: Physics

Chapter: Moment of Force

Answer 2

Answer:

Counterclockwise

Explanation:

We need to calculate the clocwkise and the counterclockwise torque.

The clockwise torque is the one generated by the child sitting on the right. This torque is given by:

$M_C = F d$

where

$F=mg=(21 kg)(9.8 m/s^2)=205.8 N$ is the force exerted by the child (his weight)

d = 1.8 m is the distance from the pivot point

So, the clockwise torque is

$M_C = (205.8 N)(1.8 m)=370.4 Nm$

The counterclockwise torque is the one generated by the adult pushing on the left, and it is given by

$M_A = F d$

where

F = 151 N is the force applied

d = 3.0 m is the distance from the pivot

Substituting,

$M_A = (151 N)(3.0 m)=453 Nm$

So, the net torque is

$M_A - M_C = 453 Nm - 370.4 Nm=82.6 Nm$

And since the counterclockwise momentum is greater than the clockwise one, the teeter-totter will rotate counterclockwise.